tìm x, biết :
\(134-2\left\{156-6\left[54-2\left(9+6\right)\right]\right\}.x=86\)
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134 - 2 . {156 - 6[54 - 2(9 + 6)]} . x = 86
134 - 2 . { 156 - 6 [ 54 - 2 . 15 ) ] } . x = 86
134 - 2 . { 156 - 6 [ 54 - 30 ] } . x = 86
134 - 2 . { 156 - 6 . 24 } . x = 86
134 - 2 . { 156 - 144 } . x = 86
134 - 2 . 21 . x = 86
2 . 21 . x = 134 - 86
2 . 21 . x = 48
21 . x = 48 : 2
21 . x = 24
x = 24 : 21
x = 1
a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)
=>x-300=0
hay x=300
\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x^2-x-6x-6\)
\(\Leftrightarrow18x+16=x^2-7x-6\)
\(\Leftrightarrow x^2-7x-18x=16+6\)
\(\Leftrightarrow x^2-15x=22\)
\(\Leftrightarrow x^2-15x-22=0\)
......
\(\Leftrightarrow\left(6x^2+27x+4x+18\right)-\left(6x^2+x+12x+2\right)=x-1-x+6\)
\(\Leftrightarrow6x^2+31x+18-6x^2-x-12x-2=7\)
\(\Leftrightarrow18x+16=7\)
\(\Leftrightarrow18x=-9\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(3x\left(2x+9\right)+2\left(2x+9\right)-x\left(6x+1\right)-2\left(6x+1\right)=x-1-x+6\)
\(6x^2+27x+4x+18-6x^2-x-12x-2=5\)
\(6x^2+\left(27x+4x\right)+18-6x^2-\left(12x+x\right)-2=5\)
\(6x^2+31x+18-6x^2-13x-2=5\)
\(\left(6x^2-6x^2\right)+\left(31x-13x\right)+\left(18-2\right)=5\)
\(18x+16=5\)
\(18x=5+16\)
\(18x=21\)
\(x=21:18\)
\(x=\frac{7}{6}\)
Vậy \(x=\frac{7}{6}\)
P/s: Mình mới lớp 6 nên hi vọng bn xem bài của mik thật kĩ xem có sai sót không,cảm ơn.
a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1-3x2)=54
\(\Rightarrow\)x3+9x2+27x+27-x(9x2+6x+1)+(2x+1)(x2-2x+1)=54
\(\Rightarrow\)x3+9x2+27x+27-9x3-6x2-x+2x3-4x2+2x+x2-2x+1=54
\(\Rightarrow\)-6x3+26x+28=54
\(\Rightarrow\)-6x3+26x=54-28
\(\Rightarrow\)-6x3+26x=26
\(\Rightarrow\)-6x3+26x-26=0
\(\Rightarrow\)-2(3x3+13x+14)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
<=>(x3-9x2+27x-27)-(x3-33+6(x2+2x+1)=15
<=>x3-9x2+27x-27-x3+27+6x2+12x+6=15
<=>-3x2+39x+9=0
<=>x2-13x+3=0
<=>(x2-2.x.13/2+169/4)-157/4=0
<=>(x-13/2)2=157/4
<=>x-13/2=\(\sqrt{\frac{157}{2}}\)hoặc x=13/2= - \(\sqrt{\frac{157}{2}}\)
<=>x=(13+\(\sqrt{\frac{157}{2}}\))hoặc x=\(\frac{13-\sqrt{\frac{157}{2}}}{2}\)
(x-3)3 - (x-3)(x2+3x+9) + 6(x+1)2 = 15
x3 -9x2 + 27x - 27 - (x3-27) + 6( x2+ 2x + 1) =15
x3 -9x2 + 27x - 27 - x3+ 27 + 6x2 + 12x + 6 = 15
-3x2 + 39x -9 = 0
-3(x2 - 13x + 3) = 0
x2 - 13x + 3 = 0
=> x=0,2350179569 ( chỗ này bấm máy tính)
còn giải thì làm theo mấy cách trong đây
BÀI 3 – 4
Phương trình bậc hai một ẩn – Công thức nghiệm
ko viết lại đề
=> \(2x\left\{156-6\left[54-2.15\right]\right\}=134-86\)
=> \(2x\left\{156-6\left[54-30\right]\right\}=48\)
=> \(2x\left\{156-6.24\right\}=48\)
=> \(2x\left\{156-144\right\}=48\)
=> \(2x.12=48\)
=> \(24x=48\)
=> \(x=48:24=2\)
Vậy \(x\in\left\{2\right\}\)